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Nadya [2.5K]
3 years ago
8

Refer to the following frequency distribution of days absent during a calendar year by employees of a manufacturing company:

Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

(a) 91 employees were absent fewer than six days.

(b) 22 employees were absent more than five days.

(c) 20 employees were absent from 6 up to 12 days.

Step-by-step explanation:

The data for the number of days absent during a calendar year by employees of a manufacturing company is given below.

(a)

The number of employees that were absent for fewer than six days is =

Frequency\  for\  class\ [0\ - \ 3]+Frequency\  for\  class\ [3\ - \ 6]\\=60 +31\\=91

Thus, there were 91 employees who were absent for fewer than six days.

(b)

The number of employees that were absent for more than 5 days is =

Frequency\  for\  class\ [6\ -\ 9]+Frequency\  for\  class\ [9\ -\12]+\\Frequency\ for\  class\ [12\ - \15]\\=14+6+2\\=22

Thus, there were 22 employees who were absent for more than 5 days.

(c)

The number of employees that were absent from 6 up to 12 days is =

Frequency\  for\  class\ [6\ -\ 9]+Frequency\  for\  class\ [9\ -\12]=14+6\\=20

Thus, there were 20 employees who were absent from 6 up to 12 days.

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yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

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P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

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=(0.0009765625)(462)

=0.451171875\approx0.4512

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Answer:

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26=100x

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